1. **Problem statement:** Prove that for all natural numbers $a$ and $b$, the equation $$(a^2 + b^2)^2 - (a^2 - b^2)^2 = (2ab)^2$$ holds true. 2. **Formula and rules:** This problem involves expanding binomials and using algebraic identities. Recall the difference of squares formula: $$x^2 - y^2 = (x - y)(x + y)$$. 3. **Step 1: Apply the difference of squares to the left side:** $$ (a^2 + b^2)^2 - (a^2 - b^2)^2 = \big((a^2 + b^2) - (a^2 - b^2)\big) \times \big((a^2 + b^2) + (a^2 - b^2)\big) $$ 4. **Step 2: Simplify each factor:** $$ \big((a^2 + b^2) - (a^2 - b^2)\big) = a^2 + b^2 - a^2 + b^2 = 2b^2 $$ $$ \big((a^2 + b^2) + (a^2 - b^2)\big) = a^2 + b^2 + a^2 - b^2 = 2a^2 $$ 5. **Step 3: Multiply the simplified factors:** $$ (2b^2)(2a^2) = 4a^2b^2 $$ 6. **Step 4: Simplify the right side:** $$ (2ab)^2 = 4a^2b^2 $$ 7. **Step 5: Conclusion:** Since both sides equal $4a^2b^2$, the equation is true for all natural numbers $a$ and $b$. **Final answer:** $$(a^2 + b^2)^2 - (a^2 - b^2)^2 = (2ab)^2$$ is proven true.